Sum of all the prime divisors of a number
Given a number N. The task is to find the sum of all the prime divisors of N.
Examples:
Input: 60 Output: 10 2, 3, 5 are prime divisors of 60 Input: 39 Output: 16 3, 13 are prime divisors of 39
A naive approach will be to iterate for all numbers till N and check if the number divides N. If the number divides N, check if that number is prime or not. Add all the prime numbers till N which divides N.
Below is the implementation of the above approach:
C++
// CPP program to find sum of // prime divisors of N #include <bits/stdc++.h> using namespace std; #define N 1000005 // Function to check if the // number is prime or not. bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that we can skip // middle five numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // function to find sum of prime // divisors of N int SumOfPrimeDivisors(int n) { int sum = 0; for (int i = 1; i <= n; i++) { if (n % i == 0) { if (isPrime(i)) sum += i; } } return sum; } // Driver code int main() { int n = 60; cout << "Sum of prime divisors of 60 is " << SumOfPrimeDivisors(n) << endl; } |
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Java
// Java program to find sum // of prime divisors of N import java.io.*; import java.util.*; class GFG { // Function to check if the // number is prime or not. static boolean isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that // we can skip middle five // numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // function to find // sum of prime // divisors of N static int SumOfPrimeDivisors(int n) { int sum = 0; for (int i = 1; i <= n; i++) { if (n % i == 0) { if (isPrime(i)) sum += i; } } return sum; } // Driver code public static void main(String args[]) { int n = 60; System.out.print("Sum of prime divisors of 60 is " + SumOfPrimeDivisors(n) + "\n"); } } |
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C#
// C# program to find sum // of prime divisors of N using System; class GFG { // Function to check if the // number is prime or not. static bool isPrime(int n) { // Corner cases if (n <= 1) return false; if (n <= 3) return true; // This is checked so that // we can skip middle five // numbers in below loop if (n % 2 == 0 || n % 3 == 0) return false; for (int i = 5; i * i <= n; i = i + 6) if (n % i == 0 || n % (i + 2) == 0) return false; return true; } // function to find // sum of prime // divisors of N static int SumOfPrimeDivisors(int n) { int sum = 0; for (int i = 1; i <= n; i++) { if (n % i == 0) { if (isPrime(i)) sum += i; } } return sum; } // Driver code public static void Main() { int n = 60; Console.WriteLine("Sum of prime divisors of 60 is " + SumOfPrimeDivisors(n) + "\n"); } } // This code is contributed // by inder_verma. |
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Python 3
# Python 3 program to find # sum of prime divisors of N N = 1000005 # Function to check if the # number is prime or not. def isPrime(n): # Corner cases if n <= 1: return False if n <= 3: return True # This is checked so that # we can skip middle five # numbers in below loop if n % 2 == 0 or n % 3 == 0: return False i = 5 while i * i <= n: if (n % i == 0 or n % (i + 2) == 0): return False i = i + 6 return True # function to find sum # of prime divisors of N def SumOfPrimeDivisors(n): sum = 0 for i in range(1, n + 1) : if n % i == 0 : if isPrime(i): sum += i return sum # Driver code n = 60print("Sum of prime divisors of 60 is " + str(SumOfPrimeDivisors(n))) # This code is contributed # by ChitraNayal |
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PHP
<?php // PHP program to find sum // of prime divisors of N $N = 1000005; // Function to check if the // number is prime or not. function isPrime($n) { global $N; // Corner cases if ($n <= 1) return false; if ($n <= 3) return true; // This is checked so that // we can skip middle five // numbers in below loop if ($n % 2 == 0 || $n % 3 == 0) return false; for ($i = 5; $i * $i <= $n; $i = $i + 6) if ($n % $i == 0 || $n % ($i + 2) == 0) return false; return true; } // function to find sum // of prime divisors of N function SumOfPrimeDivisors($n) { $sum = 0; for ($i = 1; $i <= $n; $i++) { if ($n % $i == 0) { if (isPrime($i)) $sum += $i; } } return $sum; } // Driver code $n = 60; echo "Sum of prime divisors of 60 is " . SumOfPrimeDivisors($n); // This code is contributed // by ChitraNayal ?> |
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Output:
Sum of prime divisors of 60 is 10
Time Complexity: O(N * sqrt(N))
Efficient Approach : The complexity can be reduced using Sieve of Eratosthenes with some modifications. The modifications are as follows:
- Take an array of size N and substitute zero in all the indexes(initially consider all the numbers are prime).
- Iterate for all the numbers whose indexes have zero(i.e., it is prime numbers).
- Add this number to all it’s multiples less than N
- Return the array[N] value which has the sum stored in it.
Below is the implementation of the above approach.
C++
// CPP program to find prime divisors of // all numbers from 1 to n #include <bits/stdc++.h> using namespace std; // function to find prime divisors of // all numbers from 1 to n int Sum(int N) { int SumOfPrimeDivisors[N+1] = { 0 }; for (int i = 2; i <= N; ++i) { // if the number is prime if (!SumOfPrimeDivisors[i]) { // add this prime to all it's multiples for (int j = i; j <= N; j += i) { SumOfPrimeDivisors[j] += i; } } } return SumOfPrimeDivisors[N]; } // Driver code int main() { int N = 60; cout << "Sum of prime divisors of 60 is " << Sum(N) << endl; } |
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Java
// Java program to find // prime divisors of // all numbers from 1 to n import java.io.*; import java.util.*; class GFG { // function to find prime // divisors of all numbers // from 1 to n static int Sum(int N) { int SumOfPrimeDivisors[] = new int[N + 1]; for (int i = 2; i <= N; ++i) { // if the number is prime if (SumOfPrimeDivisors[i] == 0) { // add this prime to // all it's multiples for (int j = i; j <= N; j += i) { SumOfPrimeDivisors[j] += i; } } } return SumOfPrimeDivisors[N]; } // Driver code public static void main(String args[]) { int N = 60; System.out.print("Sum of prime " + "divisors of 60 is " + Sum(N) + "\n"); } } |
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Python3
# Python 3 program to find # prime divisors of # all numbers from 1 to n # function to find prime # divisors of all numbers # from 1 to n def Sum(N): SumOfPrimeDivisors = [0] * (N + 1) for i in range(2, N + 1) : # if the number is prime if (SumOfPrimeDivisors[i] == 0) : # add this prime to # all it's multiples for j in range(i, N + 1, i) : SumOfPrimeDivisors[j] += i return SumOfPrimeDivisors[N] # Driver code N = 60print("Sum of prime" , "divisors of 60 is", Sum(N)); # This code is contributed # by Smitha |
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C#
// C# program to find // prime divisors of // all numbers from 1 to n using System; class GFG { // function to find prime // divisors of all numbers // from 1 to n static int Sum(int N) { int []SumOfPrimeDivisors = new int[N + 1]; for (int i = 2; i <= N; ++i) { // if the number is prime if (SumOfPrimeDivisors[i] == 0) { // add this prime to // all it's multiples for (int j = i; j <= N; j += i) { SumOfPrimeDivisors[j] += i; } } } return SumOfPrimeDivisors[N]; } // Driver code public static void Main() { int N = 60; Console.Write("Sum of prime " + "divisors of 60 is " + Sum(N) + "\n"); } } // This code is contributed // by Smitha |
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PHP
<?php // PHP program to find prime // divisors of all numbers // from 1 to n // function to find prime // divisors of all numbers // from 1 to n function Sum($N) { for($i = 0; $i <= $N; $i++) $SumOfPrimeDivisors[$i] = 0; for ($i = 2; $i <= $N; ++$i) { // if the number is prime if (!$SumOfPrimeDivisors[$i]) { // add this prime to // all it's multiples for ($j = $i; $j <= $N; $j += $i) { $SumOfPrimeDivisors[$j] += $i; } } } return $SumOfPrimeDivisors[$N]; } // Driver code $N = 60; echo "Sum of prime divisors of 60 is " . Sum($N); // This code is contributed by Mahadev99 ?> |
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Output:
Sum of prime divisors of 60 is 10
Time Complexity: O(N * log N)
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